Respuesta :
Answer:
30.15% of the cars are going faster than 75 mph.
Step-by-step explanation:
We are given that Speeds of automobiles on a certain stretch of freeway at 11:00 PM are normally distributed with mean 65 mph.
Also, Twenty percent of the cars are traveling at speeds between 55 and 65 mph.
Let X = Speeds of automobiles on a certain stretch of freeway
So, X ~ Normal([tex]\mu=65,\sigma^{2} = \sigma^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean speed of automobiles = 65 mph
[tex]\sigma[/tex] = standard deviation
Now, we are given that Twenty percent of the cars are traveling at speeds between 55 and 65 mph, that means;
P(55 mph < X < 65 mph) = 0.20
P(55 mph < X < 65 mph) = P(X < 65 mph) - P(X [tex]\leq[/tex] 55 mph)
P(X < 65 mph) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{65-65}{\sigma}[/tex] ) = P(Z < 0) = 0.50 {using z table}
P(X [tex]\leq[/tex] 65 mph) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{55-65}{\sigma}[/tex] ) = P(Z [tex]\leq[/tex] [tex]\frac{-10}{\sigma}[/tex] ) = 1 - P(Z < [tex]\frac{10}{\sigma}[/tex] )
SO, P(55 mph < X < 65 mph) = 0.20
P(X < 65 mph) - P(X [tex]\leq[/tex] 55 mph) = 0.20
0.50 - [1 - P(Z < [tex]\frac{10}{\sigma}[/tex] ) ] = 0.20
P(Z < [tex]\frac{10}{\sigma}[/tex] ) - 0.50 = 0.20
P(Z < [tex]\frac{10}{\sigma}[/tex] ) = 0.70
Now, looking at the z table we find that the critical value of x which gives a probability area less than 70% is 0.5244, that means;
[tex]\frac{10}{\sigma}[/tex] = 0.5244
[tex]\sigma=\frac{10}{0.5244}[/tex] = 19.07 mph
Therefore, standard deviation, [tex]\sigma[/tex] = 19.07 mph.
Now, percentage of the cars that are going faster than 75 mph is given by = P(X > 75 mph)
P(X > 75 mph) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{75-65}{19.07}[/tex] ) = P(Z > 0.52) = 1 - P(Z [tex]\leq[/tex] 0.52)
= 1 - 0.6985 = 0.3015
Therefore, 30.15% of the cars are going faster than 75 mph.
